Transcript Department of Computer and Information Science, School of Science, IUPUI CSCI 230 Arrays Case Study Dale Roberts, Lecturer IUPUI [email protected] Dale Roberts.

Department of Computer and Information Science,
School of Science, IUPUI
CSCI 230
Arrays
Case Study
Dale Roberts, Lecturer
IUPUI
[email protected]
Dale Roberts
Sorting Arrays
Sorting data
Important computing application
Virtually every organization must sort some data
Bubble sort (sinking sort)
Several passes through the array
Successive pairs of elements are compared
If increasing order (or identical ), no change
If decreasing order, elements exchanged
Repeat
Example:
original: 3 4 2 6 7
pass 1:
3 2 4 6 7
pass 2:
2 3 4 6 7
Small elements "bubble" to the top
Dale Roberts
Searching Arrays: Linear Search vs Binary Search
Search an array for a key value
Linear search
Simple
Compare each element of array with key value
Useful for small and unsorted arrays
Binary search
For sorted arrays
Compares middle element with key
If equal, match found
If key < middle, looks in first half of array
If key > middle, looks in last half
Repeat
Very fast; at most n steps, where 2n > number of elements
30 element array takes at most 5 steps
25 > 30 so at most 5 steps
Dale Roberts
Case Study: Computing Mean, Median and
Mode Using Arrays
Mean – average
Median – number in middle of sorted list
Example: 1, 2, 3, 4, 5
 3 is the median
Mode – number that occurs most often
Example: 1, 1, 1, 2, 3, 3, 4, 5
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 1 is the mode
/* Fig. 6.16: fig06_16.c
This program introduces the topic of survey data analysis.
It computes the mean, median, and mode of the data */
#include <stdio.h>
#define SIZE 99
void
void
void
void
void
mean( const int [] );
median( int [] );
mode( int [], const int [] ) ;
bubbleSort( int [] );
printArray( const int [] );
Function prototypes
int main()
{
int frequency[ 10 ] = { 0 };
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int response[
{ 6, 7, 8,
7, 8, 9,
6, 7, 8,
7, 8, 9,
6, 7, 8,
7, 8, 9,
5, 6, 7,
7, 8, 9,
7, 4, 4,
4, 5, 6,
SIZE ] =
9, 8, 7,
5, 9, 8,
9, 3, 9,
8, 9, 8,
7, 8, 7,
8, 9, 8,
2, 5, 3,
6, 8, 7,
2, 5, 3,
1, 6, 5,
8,
7,
8,
9,
9,
9,
9,
8,
8,
7,
9,
8,
7,
7,
8,
7,
4,
9,
7,
8,
8, 9,
7, 8,
8, 7,
8, 9,
9, 2,
5, 3,
6, 4,
7, 8,
5, 6,
7 };
mean( response );
median( response );
mode( frequency, response );
return 0;
Initialize array
Call functions
mean, median,
and mode
}
void mean( const int answer[] )
{
int j, total = 0;
printf( "%s\n%s\n%s\n", "********", "
Define function
mean
Mean", "********" );
for ( j = 0; j <= SIZE - 1; j++ )
total += answer[ j ];
printf( "The mean is the average value of the data\n"
"items. The mean is equal to the total of\n"
"all the data items divided by the number\n"
"of data items ( %d ). The mean value for\n"
"this run is: %d / %d = %.4f\n\n",
SIZE, total, SIZE, ( double ) total / SIZE );
}
Dale Roberts
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51 void median( int answer[] )
52 {
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printf( "\n%s\n%s\n%s\n%s",
Define function median
"********", " Median", "********",
"The unsorted array of responses is" );
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printArray( answer );
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bubbleSort( answer );
printf( "\n\nThe sorted array is" );
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printArray( answer );
printf( "\n\nThe median is element %d of\n"
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"the sorted %d element array.\n"
"For this run the median is %d\n\n",
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SIZE / 2, SIZE, answer[ SIZE / 2 ] );
65 }
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67 void mode( int freq[], const int answer[] )
68 {
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int rating, j, h, largest = 0, modeValue = 0;
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printf( "\n%s\n%s\n%s\n",
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"********", " Mode", "********" );
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for ( rating = 1; rating <= 9; rating++ )
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freq[ rating ] = 0;
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for ( j = 0; j <= SIZE - 1; j++ )
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++freq[ answer[ j ] ];
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Sort Array
Print middle element
Is this computation of the median element
correct? Notice that SIZE must be odd for
this to be correct. Is SIZE odd?
Define function mode
Notice how the subscript in freq[] is
the value of an element in response[]
(answer[])
Increase frequency[]
depending on response[]
Dale Roberts
Does this code implement the
definition of mode that you learned
in math class?
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printf( "%s%11s%19s\n\n%54s\n%54s\n\n",
"Response", "Frequency", "Histogram",
"1
1
2
2", "5
0
5
0
5" );
for ( rating = 1; rating <= 9; rating++ ) {
printf( "%8d%11d
", rating, freq[ rating ] );
if ( freq[ rating ] > largest ) {
largest = freq[ rating ];
modeValue = rating;
}
for ( h = 1; h <= freq[ rating ]; h++ )
printf( "*" );
printf( "\n" );
Print stars depending on value of
frequency[]
}
printf( "The mode is the most frequent value.\n"
"For this run the mode is %d which occurred"
" %d times.\n", modeValue, largest );
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void bubbleSort( int a[] )
{
int pass, j, hold;
for ( pass = 1; pass <= SIZE - 1; pass++ )
for ( j = 0; j <= SIZE - 2; j++ )
if ( a[
hold
a[ j
a[ j
}
j
=
]
+
Define bubbleSort
Bubble sort can make the smallest values
“sink” by scanning top to bottom using <
or make the largest values “float” by
scanning bottom to top using >.
] > a[ j + 1 ] ) {
a[ j ];
Bubble sort: if elements
= a[ j + 1 ];
out of order, swap them.
1 ] = hold;
}
void printArray( const int a[] )
{
int j;
for ( j = 0; j <= SIZE - 1; j++ ) {
if ( j % 20 == 0 )
printf( "\n" );
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Define printArray
printf( "%2d", a[ j ] );
}
129 }
Dale Roberts
Program
Output
********
Mean
********
The mean is the average value of the data
items. The mean is equal to the total of
all the data items divided by the number
of data items (99). The mean value for
this run is: 681 / 99 = 6.8788
********
Median
********
The unsorted array of responses is
7 8 9 8 7 8 9 8 9 7 8 9 5 9 8 7 8 7 8
6 7 8 9 3 9 8 7 8 7 7 8 9 8 9 8 9 7 8 9
6 7 8 7 8 7 9 8 9 2 7 8 9 8 9 8 9 7 5 3
5 6 7 2 5 3 9 4 6 4 7 8 9 6 8 7 8 9 7 8
7 4 4 2 5 3 8 7 5 6 4 5 6 1 6 5 7 8 7
The sorted
1 2 2 2 3
5 6 6 6 6
7 7 7 7 7
8 8 8 8 8
9 9 9 9 9
array
3 3 3
6 6 6
7 7 7
8 8 8
9 9 9
is
4 4
6 6
7 7
8 8
9 9
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9
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7
8
8
The median is element 49 of
the sorted 99 element array.
For this run the median is 7
********
Mode
********
Response Frequency
Histogram
5
1
0
1
5
2
0
2
5
1
1
*
2
3
***
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4
****
4
5
*****
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8
********
6
9
*********
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23
***********************
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27
***************************
9
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*******************
The mode is the most frequent value.
For this run the mode is 8 which occurred 27 times.
Dale Roberts